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Let's consider the initial value problem

$$ y^{\prime}(t) = f(t)y(t), y(0)=1 $$

where, $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, Then the IVP has

(a) infinitely many solutions for some $f$

(b) a unique solution in $\mathbb{R}$

(c) no solution in $\mathbb{R}$ for some $f$

(d) a solution in an interval containing $0$ but not on $\mathbb{R}$ for some $f$

We have to find the correct options. Can we do it like:

$$ y^{\prime}(t) = f(t)y(t)$$

$$y=c e^{\int f(t)dt}$$

$$y=c e^{g(t)}; \int f(t)dt=g(t) $$ on putting initial condition, we can find the value for $c$. In that case,can we reject options a and c. Any help, Please

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    $\begingroup$ Do you know the Picard-Lindelöf theorem? $\endgroup$ – Severin Schraven May 25 '17 at 19:44
  • $\begingroup$ No, I don't know! If it helps then please state it to explain the answer. $\endgroup$ – Hirakjyoti Das May 26 '17 at 13:14

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